
%!TEX program = xelatex
%!TEX TS-program = xelatex
%!TEX encoding = UTF-8 Unicode

\documentclass[10pt]{article} 

\input{wang_preamble.tex}

%\addtolength{\textheight}{0.5cm}
%\addtolength{\voffset}{-0.4cm}
\addtolength{\textwidth}{1.5cm}
\addtolength{\hoffset}{-0.8cm}   

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\usepackage{titling}
\setlength{\droptitle}{-2cm}   % This is your set screw

%%文档的题目、作者与日期
\author{学号 \underline{\hspace{4cm}} \hspace{1cm} 姓名 \underline{\hspace{4cm}} }
\title{实变函数测验 5A-5B}
%\date{\vspace{-3ex}}
\renewcommand{\today}{\number\year \,年 \number\month \,月 \number\day \,日}
\date{2024 年 5 月 27 日}
%\date{March 9, 2021}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{document}

\maketitle

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

圈出叙述错误的步骤，并加以改正。

\begin{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item  %Problem 01
（积分的绝对连续性）设 $E\subseteq\mathbb{R}^n$ 是可测集，设 $f\in L(E)$, 则对任意 $\varepsilon>0$, 存在 $\delta>0$, 使得对任意可测集 $A\subseteq E$, 只要 $m(A)<\delta$, 就有
\begin{subequations}
\begin{align}
\left\vert \int_A f(x)dx \right\vert < \varepsilon. 
\tag{*}
\label{main-task}
\end{align}
\end{subequations}


%证明：
\begin{enumerate}[label={(\arabic*)}]

\item  %1
因为 $f$ 是 $E$ 上的勒贝格可积函数，所以 $|f|$ 也是 $E$ 上的勒贝格可积函数。

\item  %2
根据非负简单函数的勒贝格积分的定义，对任意 $\varepsilon>0$, 存在非负简单函数 $0\le \varphi\le |f|$ 使得 
$$\int_E |f(x)|dx - \frac{\varepsilon}{2} \le \int_E\varphi(x)dx \le \int_E |f(x)|dx. $$

\item  %3
对任意可测集 $A\subseteq E$, 就有 
%\begin{eqnarray}
\begin{subequations}
\begin{align}
\left\vert \int_A f(x)dx\right\vert \le  \int_A \left\vert f(x)\right\vert dx
= \int_A \left[ \left\vert f(x)\right\vert - \varphi(x) \right] dx + \int_A\varphi(x)dx. 
\tag{**}
\label{main-estimation}
\end{align}
\end{subequations}
%\end{eqnarray}

\item  %4
根据(2)可知估计式 (\ref{main-estimation}) 最右边的第一项不超过 $\frac{\varepsilon}{2}$. 

\item  %5
记 $M=1+\max\{\varphi(x): x\in E\}$, 记 $\delta=\frac{\varepsilon}{2M}$, 
当 $m(A)\le \delta$ 时，估计式 (\ref{main-estimation}) 最右边的第二项
\begin{eqnarray*}
\int_A\varphi(x)dx < \int_A M dx = M\cdot m(A) \le M\cdot \delta = M\cdot \frac{\varepsilon}{2M} = \frac{\varepsilon}{2}. 
\end{eqnarray*}

\item  %6
将(4)与(5)代入(3)中的可知需要证明的不等式 (\ref{main-task}) 成立。


\end{enumerate}



\vspace{1.8cm}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item  %Problem 02
（法图引理）设 $E\subseteq\mathbb{R}^n$ 是可测集，设 $\{f_n(x)\}$ 是 $E$ 上一列非负可测函数，则有
$$ \int_E \varliminf_{n\to\infty} f_n(x)dx \le \varliminf_{n\to\infty} \int_E f_n(x)dx. $$ 
%举例说明等号不一定成立。

%证明：
\begin{enumerate}[label={(\arabic*)}]

\item  %1
记 $g_n(x) = \inf\{ f_n(x), f_{n+1}(x), f_{n+2}(x), \cdots \}$, 则随着 $n$ 增加，求下确界的数集里的元素越来越多，所以
\begin{subequations}
\begin{align}
g_1(x) \le g_2(x)\le \cdots \le g_n(x) \le \cdots. 
\tag{*}
\label{main-tool}
\end{align}
\end{subequations}

\item  %2
根据题设，每个 $f_n(x)$ 都是非负可测函数，所以 (\ref{main-tool}) 是非负可测函数序列。

\item  %3
根据下极限的定义、莱维定理、以及 $g_n(x)\le f_n(x)$, 可得 
\begin{eqnarray*}
\int_E \varliminf_{n\to\infty} f_n(x)dx 
= \int_E \lim_{n\to\infty} g_n(x)dx  
= \lim_{n\to\infty} \int_E g_n(x)dx 
\le \varliminf_{n\to\infty} \int_E f_n(x)dx. 
\end{eqnarray*}


\end{enumerate}



\vspace{0.3cm}

\newpage 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item  %Problem 03
（勒贝格控制收敛定理）设 $E\subseteq \mathbb{R}^n$ 是可测集，设 $\{f_n(x)\}$ 是 $E$ 上的一列收敛的可测函数，设 $F(x)$ 是 $E$ 上的非负勒贝格可积函数。设对任意的正整数 $n$, $|f_n(x)|\le F(x)$ 在 $E$  上成立，
%且 $\lim\limits_{n\to\infty} f_n(x)=f(x)$ 在 $E$ 上几乎处处成立，
则有 
\begin{subequations}
\begin{align}
\lim\limits_{n\to\infty} \int_E f_n(x) dx = \int_E \lim\limits_{n\to\infty} f_n(x) dx. 
\tag{$\heartsuit$}
\label{main-result}
\end{align}
\end{subequations}

%证明：
\begin{enumerate}[label={(\arabic*)}]

\item  %1
因为每个 $f_n(x)$ 都是可测函数，所以极限函数也是可测函数，记为
$$f(x)=\lim\limits_{n\to\infty} f_n(x). $$ 

\item  %2
因为 $|f_n(x)|\le F(x)$ 以及 $f(x)$ 是 $\{f_n(x)\}$ 的极限，所以 $|f(x)|\le F(x)$. 

\item  %3
因为 $F(x)$ 勒贝格可积，所以 $f(x)$ 和 $f_n(x)$ 都勒贝格可积，即 (\ref{main-result}) 的两边的积分都是有限的实数。

\item  %4
记 $g_n(x)=|f_n(x)-f(x)|$. 因为 $f_n(x)$ 和 $f(x)$ 都是可测函数，所以 $g_n(x)$ 是可测函数。

\item  %5
因为 $0\le g_n(x)\le 2F(x)$, 而 $2F(x)$ 是勒贝格可积的，所以 $g_n(x)$ 是勒贝格可积的。

\item  %6
对非负可测函数序列 $\{2F(x)-g_n(x)\}$ 应用法图引理，可得
%\begin{eqnarray*}
\begin{subequations}
\begin{align}
\int_E \varliminf_{n\to\infty}  [ 2F(x)-g_n(x) ] dx \le \varliminf_{n\to\infty} \int_E [2F(x)-g_n(x) ] dx. 
\tag{$\diamondsuit$}
\label{main-step-1}
\end{align}
\end{subequations}
%\end{eqnarray*}

\item  %7
因为 $g_n(x)$ 收敛于零，以及勒贝格积分的基本性质，所以 (\ref{main-step-1}) 式的左边和右边可以化为  
$$
\int_E  2F(x) dx \le \varliminf_{n\to\infty} \left[ \int_E  2F(x)dx - \int_E g_n(x) dx \right]. 
$$

\item  %8
根据数列的负数列的下极限等于原数列的上极限，所以上式可以化为   
$$
\int_E  2F(x) dx \le \int_E  2F(x)dx - \varlimsup_{n\to\infty} \int_E g_n(x) dx. 
$$

\item  %9
因为 $F(x)$ 在 $E$ 上的勒贝格积分是确定的，所以 $\int_E  2F(x) dx<\infty$, 所以由上式可得 
$$
\varlimsup_{n\to\infty} \int_E g_n(x) dx \le 0. 
$$

\item  %10
因为 $g_n(x)$ 是非负可测函数，所以 $\int_E g_n(x) dx \ge 0$. 所以由上式可得 
$$
\lim_{n\to\infty} \int_E g_n(x) dx = 0. 
$$

\item  %11
返回 $f_n(x)$ 和 $f(x)$ 的记号，上式就是
$$\lim_{n\to\infty} \int_E |f_n(x) - f(x)| dx = 0. $$

\item  %12
根据勒贝格积分的基本性质，可得 
$$
\left\vert \int_E f_n(x) dx - \int_E f(x) dx \right\vert \le 
\int_E \left\vert f_n(x) - f(x) \right\vert dx. 
$$

\item  %13
从(11)与(12)可知所需要证明的等式 (\ref{main-result}) 成立。

\end{enumerate}


\vspace{0.3cm}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{enumerate}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{document}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


